csdm - monday, october 5, 2009 - arad.pdf

The detectability lemma and quantum gap amplification

Dorit Aharonov1, Itai Arad1, Zeph Landau2 and Umesh Vazirani2

1School of Computer Science and Engineering,The Hebrew University,

Jerusalem, Israel

2Electrical Engineering and Computer Science,UC Berkeley,

California, USA

October 2, 2009

D. Aharonov, I. Arad, Z. Landau, U. Vazirani ( School of Computer Science and Engineering, The Hebrew University, Jerusalem, Israel, Electrical Engineering and Computer Science, UC Berkeley, California, USA)The Detectability & Gap amplification October 2, 2009 1 / 24

Quantum Hamiltonian complexity

Settings:n-variables,local constraints

Main concepts:Satisfying assignments,Reductions,Gap amplification,PCP

CSP

Settings:n quantum particleslocal quantum constrains (Hamiltonians)

Main concepts:“Satisfying” quantum states (ground states)Structure of entanglement

Condensed Matter

Hamiltonian ComplexityMain result: Quantum Cook-Levin (Kitaev ’02)

Analog of CSP results for Hamiltonians:Reductions,

completeness,(PCP ?)

Classical simulationsof Hamiltonian

systems


CSP in quantum languageMAX-3-SATSettings:

n bits: x1, . . . xn

M constraints:f(x1, . . . , xn) = C1 ∧ · · · ∧ CMCi – 3-local CNF clause

Goal: find the minimal possible # ofviolations

MAX-3-SAT – in quantum languageSettings:

Hilbert space of n qubits – 2n basisstates:|00 · · · 00〉 , |00 · · · 01〉 , . . . , |11 · · · 11〉Hamiltonian with M terms:H = Q1 + . . .+QM

Qi – 3-local “classical projections”

Goal: approximate the lowest eigenvalueof H: λ(H) = 0 or λ(H) ≥ 1.

Example:Ci = (xk ∨ x` ∨ xm)

Violating when (xk, x`, xm) = (0, 1, 0)

=⇒

00

10

00

00

|010〉

|010〉

0

0Qi = ⊗1

CSP is satisfiable ⇔ λ(H) = 0


Moving to eigenvectors and eigenvalues

Local Hamiltonian: H = Q1 +Q2 + . . . 2n × 2n matrix

Ci = (xk ∨ x` ∨ xm)

Violating when (xk, x`, xm) = (0, 1, 0)=⇒

00

10

00

00

|010〉

|010〉

0

0Qi = ⊗1

Eigenvectors: H |x1, . . . , xn〉 = (# violations) |x1, . . . , xn〉

Eigenvalue of |x1, . . . , xn〉: # of violations.

λ(H) = Lowest eigenvalue = minimal # of violations.

The problem: Decide whether λ(H) = 0 or λ(H) ≥ 1


Moving to the general quantum CSP

Quantum CSP (The Local Hamiltonian problem)

H =PMi=1Qi general local projections λ(H) = lowest eigenvalue of H

Decide: λ(H) = 0 or λ(H) ≥ 1poly(n)

Note:Qi are no longer diagonal in the standard basis,and the eigenvectors of H are superpositions:

|ψ〉 =

2n termsz | a1 |0 · · · 00〉+ a2 |0 · · · 01〉+ . . .

A different view on LH:

λ(H) = minψ〈ψ|H |ψ〉

= minψ〈ψ|

Xi

Qi |ψ〉

= minψ

Xi

〈ψ|Qi |ψ〉

0 ≤ 〈ψ|Qi |ψ〉 ≤ 1 –The energy of the constraint:how much it is violated.

(compare to 0 or 1 in the classical case)


Hamiltonian Complexity - The quantum analog of CSP

Quantum NP (QMA)

Decision problems that can be solved with a quantumwitness |ψ〉 and a polynomial quantum verifier Vx.

x ∈ L : ∃ |ψx〉 s.t. Pr[V (x, |ψx〉) = yes] ≥ 2/3 ,

x /∈ L : ∀ |φ〉 ,Pr[V (x, |φ〉) = yes] ≤ 1/3 .

x |ψx〉

V

NOYES

Quantum Cook-LevinThe Local-Hamiltonian problem is QMA-complete (Kitaev, 98).

Inclusion is the easy direction

Hardness is similar to Cook-Levin – but with some quantum twists.


Central results in Hamiltonian Complexity

Many results and techniques can be imported from CSP to Hamiltonian complexity:

Reductions using Gadgets

2-Local (even planar) Hamiltonian is QMAcomplete

Satisfaction Threshold

Lovasz local lemma

Aharonov, van Dam, Kempe, Landau,Lloyd, Regev ’04Kempe, Kitaev, Regev ’04Oliveira, Terhal ’05Najag ’07Shondhi ’09Bravyi ’09Ambainis, Kempe, Sattath ’09

But sometimes things are different:

Local Hamiltonian in 1d is QMA-complete (but1d CSP is in P).

Aharonov, Gottesman, Irani, Kempe’07

What about Quantum PCP (QPCP) ?


Quantum PCP

Two ways to formulate the QPCP conjecture:

QPCP conjecture I∀L ∈ QMA there is a quantum

verifier that has access to only q

random qubits from the witness.

Quantumreduction⇐⇒

QPCP conjecture IIDeciding whether

λ(H) = 0 or

λ(H) ≥ c ·Mis QMA-hard.

This is, however, a difficult problem


Difficulties in proving QPCP

Main Problem: no cloningThere is no quantum transformation |ψ〉 |0〉 7→ |ψ〉 |ψ〉 for all |ψ〉

No cloning was thought to prevent QECCs – but QECCs exist!

QPCP seems to require even more (QECCs are not locally decodable)

Possible implications of resolving this question:


We focus on a central ingredient in Dinur’s PCP proof, which does not require cloning:

Gap Amplification


What we want to amplify

In the Classical case

UNSATdef=

min # violationsM

Find an efficient transformation to anew CSP s.t.,

0

1

p0

1c · t · p

UNSAT

Factor t

amp’

In the Quantum case

QUNSATdef=

λ(H)

M

Find an efficient transformation to anew L.H. s.t.,

0

1

p0

1c · t · p

QUNSAT

Factor t

amp’


The probability of detecting a violation in a random constraint

Classically:

M constraints

pMviolations

p – Probability of picking a violatedconstraint

p =# of violations

M≥ UNSAT

Quantumly:

Quantum Measurement

|ψ〉Π|ψ〉

(1− Π)|ψ〉

Prob = ‖Πψ‖2

Prob = ‖(1− Π)ψ‖2

1

0

In our case, when measuring Qi:

Pr[1] = ‖Qi |ψ〉 ‖2 = 〈ψ|Qi |ψ〉= constraint’s energy

The probability of measuring a violation fora random i.

p =1

M

Xi

〈ψ|Qi |ψ〉 =λ(H)

M≥ QUNSAT

We want to amplify the probability of measuring a violation


Amplification by repitition

Classically

Pick t random constraints

Probability of at least 1 violation:

Pr = 1− (1− p)t ' t · p

New system size: M t

M constraints

pMviolations

Quantumly, this also works:

Choose t random constraintsMeasure them one after the otherAfter every measurement, the system collapses into a new state:|ψ1〉 → |ψ2〉 → · · · → |ψt〉, but always:P

i 〈ψj |Qi |ψj〉M

≥ λ(H)

M

Probability of measuring at least one violation:

Pr ≥ 1− (1− λ(H)/M)2 ' t · λ(H)/M


Amplification with expanders

Ajtai et al, Impagliazzo & Zuckerman – RP & BPP amp’

Expander graphsTaking a t-walk on a d-regular expander graph isalmost like picking t random edges.Advantage: The new system is much smaller: Mdt

(instead of M t)

satisfiededges

unsatisfiededges

C1

C2 C3 C4C5

Ce = C1 ∧ C2 ∧ C3 ∧ C4 ∧ C5


Quantum Conspiration?

Conspiration Theory

Classical correlations decay exponentially fast on anexpander, but is this also true for quantum correlations?

How do we know that the expander is “random enough”?– Can the system trick us by collapsing adversely ?

Classically: well-defined partition.

Quantumly:no assignment⇒ no partition.

Quantum World

Energy distribution

Classical World

satisfiededges

unsatisfiededges

Is there a distribution of assignments? For two constraints A,B, Pr(A = 1, B = 0)is not well defined:

Pr

h(A=1) → (B=0)

i= ‖(I −B)A |ψ〉 ‖2 6= ‖A(I −B) |ψ〉 ‖2 = Pr

h(B=0) → (A=1)

iA and B do not commute⇒ no distribution over assignments


Our Results

Quantum Gap AmplificationConsider a quantum CSP problem on a d-regular expander graph G = (V,E) with asecond largest eigvenvalue 0 < λ < 1. The edges are projections and the vertices arequdits of dimension W .

A1

A2 A3A4A5

A~e = A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5

Let Gt be the hyper graph of t-walks derived from G, and for each such t-walk define aprojection into the intersection of all accepting subspaces along the walk.

Then:

QUNSAT(Gt) ≥ c(λ)K(q, d) min“t ·QUNSAT(G), 1

”D. Aharonov, I. Arad, Z. Landau, U. Vazirani ( School of Computer Science and Engineering, The Hebrew University, Jerusalem, Israel, Electrical Engineering and Computer Science, UC Berkeley, California, USA)The Detectability & Gap amplification October 2, 2009 16 / 24

Layers

layer 1layer 2layer 3layer 4

Typically, constraints can be arranged in a finite number g of layers.

In a fixed layer, all projections commute and can be measured simultaneously –there exists an underlying joint probability distribution.

With respect to one layer, we have a distribution of classical systems:

= + + + . . .

We can use the classical PCP result on each member.

There are only g layers⇒ there must be a layer with a constant energy.

But there’s a problem . . .


The problem (yet another conspiration)

It is trivial to satisfy all constraints in one fixed layer.But not simultaneously in all layers (because λ(H) > 0).

What if the violation/satisfation distribution in every layer conspires to be:

Violations

Weight

1− 1/poly(n)

1/poly(n)

0% 100%

|ψ〉 = 0.99 |no violations〉+ 0.01 |full violation〉

No amplification:In the “no violations” part there isno amplification because allconstraints are satisfied.

In the “full violations” part there isno amplification because we havereached the maximum.

We must somehow rule out this possibility if we want to show amplification


The detectability lemma

The detectability lemma (` = 0 case)Settings:⇒ Π(j) is the projection into the satisfying subspace of the j’th layer⇒ Minimal energy is ε0

def= λ(H) > 0

⇒ Projections are taken from a fixed set.

Then:‖Π(1) · · ·Π(g) |ψ〉 ‖2 ≤ 1

ε0/c+ 1' 1− ε0/c

where c is constant independent of n.

In a sequential measurent of all layers, the probability of not detecting any violations isbounded away from 1 by a constant.

CorollaryThere must be at least one layer j with a constant projection on the violating subspace

‖(I −Π(j)) |ψ〉 ‖ ≥ c′


General outline of the proofDefine

|Ω〉 def= Π(1) · · ·Π(g) |ψ〉

We wish to show ‖Ω‖2 ≤ 1ε0/c+1

.Note:

‖Ω‖2ε0 ≤ 〈Ω|H |Ω〉

We prove:

〈Ω|H |Ω〉 ≤ c`1− ‖Ω‖2

´⇓

‖Ω‖2ε0 ≤ c`1− ‖Ω‖2

´⇓

‖Ω‖2 ≤ 1

ε0/c+ 1

In the commuting case, 〈Ω|H |Ω〉 = 0 (trivial)In the general case, we show that energy contribution due to non-commuteness isexponentially decreasing.We must find a way to quantify non-commuteness.


The XY decomposition I – Pyramids

Pyramids

For every projection Q, we definea set of projections Pyr[Q]

Hpyr - The supporting Hilbertspace.

Hpyr︷︸︸︷Q

The XY decomposition

Hpyr = X ⊕ YX = common eigenvectors

Y = The rest

PX , PY = Projections to X,Y

Observations:PX , PY commute with all Q′ ∈ Pyr[Q]

‖PY`Q

Q′∈Pyr[Q]Q′´PY ‖ ≤ θ < 1

If the projections are drawn from afixed set, θ < 1 is constant!


The XY decomposition II – PonzisA Ponzi is a set of constraints from a fixed layer, whose pyramids don’t intersect

H(1)pyr︷︸︸︷ H(2)

pyr︷︸︸︷ H(3)pyr︷︸︸︷

A constant number of Ponzi’s is needed to cover all constraintsEponz – The energy operator of a Ponzi.The entire Hilbert space is decomposed in terms of XY sectors ν:

An XY sector: ν = (X,X, Y,X, Y, . . .)

Projection onto ν: Pν = PX ⊗ PX ⊗ PY ⊗ PX · · ·|ν| = No. of Y spaces

|Ω〉 =Xν

Pν |Ω〉def=Xν

|Ων〉

Eponz commutes with the decomp’:

〈Ω|Eponz |Ω〉 =Xν

〈Ων |Eponz |Ων〉


Open questions

The quantum amplification lemma is trivial to prove when all the projectionscommute (one does not need the detectability lemma). Still it is not clear:

I if the quantum PCP can be proved for commuting HamiltoniansI what is the complexity of this special class? is it NP? QMA ?

Can the XY decomposition or the detectability lemma be used elsewhere tohandle the non-commuteness of the LH problem ?

Are there any interesting implications of the detectability lemma and the XYdecomposition to solid-state physics?

Currently, the detectability lemma allows us an RP type of amplification (one-sidederrors). Can it be generalized to prove a BPP type of amplification (two-sidederrors) ?

What is the right definition of quantum PCP? (one-sided errors/two sided errors?)

Can we prove a quantum PCP with exponential witness ?

If there is no quantum PCP theorem, then what is the complexity of approximatingQUNSAT(H) up to a constant? It must be NP-hard - but is it inside NP?


csdm - monday, october 5, 2009 - arad.pdf

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